Problem: A herd of dinosaurs made paintings in the sand with their claws. Each baby dinosaur made $15$ paintings and each adult dinosaur made $7$ paintings. The entire herd made $208$ paintings in total, and there were $3$ times as many baby dinosaurs as adult dinosaurs. How many baby dinosaurs and adult dinosaurs were there? There were
Explanation: Let $x$ represent the number of baby dinosaurs and let $y$ represent the number of adult dinosaurs. Since we have two unknowns, we need two equations to find them. Let's use the given information to write two equations containing $x$ and $y$. For instance, we know that each baby dinosaur made $\textit{15}$ paintings, each adult dinosaur made $\textit{7}$ paintings, and the total number of paintings made was $\textit{208}$. How can we model this sentence algebraically? The total number of paintings made by baby dinosaurs can be modeled by $15x$, and the total number of paintings made by adult dinosaurs can be modeled by $7y$. Since these together add up to $208$, we get the following equation: $15 x+7 y = 208$ We are also given that there were $\textit{3}$ times as many baby dinosaurs as adult dinosaurs. This can be expressed as: $x = 3 y $ Now that we have a system of two equations, we can go ahead and solve it! Let's substitute $ x={3 y}$ into the first equation: $\begin{aligned}15 x+7 y &= 208\\\\ 15 \cdot ({3y})+7y&=208\\\\ 45 y+7y &=208\\\\ y&=4\end{aligned}$ Now we can substitute $y = 4$ into $x=3 y$ and find that $x=12$. Recall that $x$ denotes the number of baby dinosaurs and $y$ denotes the number of adult dinosaurs. Therefore, there were $\textit{12}$ baby dinosaurs and $\textit{4}$ adult dinosaurs.